Family of Lines MCQ Quiz in తెలుగు - Objective Question with Answer for Family of Lines - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Given line is \(px^{2} - qy^{2} = 0\)

General equation is \(ax^{2}+by^{2}+h=0\)

Comparing above equation with \((i)\), we get

\(a = p, b = -q, h = 0\)

Lines are real and distinct if \(h^{2} - ab > 0\)

\(\Rightarrow 0 + pq > 0\)

\(\implies pq > 0\)

(A) \( x^{2} - x = 0 \)

\( x^{2} - x = 0 \)

\( x(x-1) = 0 \)

\( \implies x = 0 \) and \( x = 1 \)

So, it represents a pair of lines \( x = 0 \) and \( x = 1 \)

(B) \( xy - x = 0 \)

\( x(y-1) = 0 \)

\( \implies x = 0 \) and \( y = 1 \)

So, it represents a pair of lines \( x = 0 \) and \( y = 1 \)

(C) \( y^{2} - x + 1 = 0 \)

\( \implies y^{2} = x - 1 \) which represents parabola

(D) \( xy + x + y + 1 = 0 \)

\( \implies x(y+1) + y + 1 = 0 \)

\( \implies (y+1)(x+1) = 0 \)

\( \implies y = -1 \) and \( x = -1 \)

So, it represents a pair of lines \( x = -1 \) and \( y = -1 \)

Hence, only (C) does not represents a pair of lines.

Calculation

4a2 + 9b2 – c2 + 12ab = 0
⇒ (2a + 3b)² − c² = 0

⇒ 2a + 3b - c = 0 or 2a + 3b + c = 0

⇒ c = ±(2a + 3b)

∴ ax + by + c = 0

⇒ ax + by ± (2a + 3b) = 0

⇒ a(x ± 2) + b(y ± 3) = 0 (family of lines)

⇒ (-2, -3) or (2, 3)

Hence option 1 is correct 

Calculation

\(\rm r^2\cos^2\left(\theta-\frac{\pi}{3}\right)=2\)

⇒ r²(cosθ.cos\(\frac{\pi}{3}\) ​+ sinθ.sin\(\frac{\pi}{3}\) )² = 2

⇒ \(r^2 (\frac{cosθ}{2}+\frac{\sqrt{3}}{2}sinθ)^{2} = 2\)

⇒ (rcosθ + √3rsinθ)² = 8

⇒ (x + y√3)² = 8

⇒ (x + y√3 - 2√2)(x + y√3 + 2√2) = 0

Represents a pair of straight lines

Hence option 4 is correct

Calculation

Locus of point P(x, y) whose distance from  x + 2y + 7 = 0 & 2x – y + 8 = 0 are equal is 

⇒ \(\frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}\)

⇒ (x + 2y + 7)² – (2x – y + 8)² = 0 

Combined equation of lines

⇒ (x – 3y + 1) (3x + y + 15) = 0   { \(\because \) a² - b² = (a + b)(a -  b)}

⇒ 3x² – 3y² – 8xy + 18x – 44y + 15 = 0 

⇒ \(x^2-y^2-\frac{8}{3} x y+6 x-\frac{44}{3} y+5=0\)

⇒ x² – y² + 2h xy + 2gx 2 + 2fy + c = 0 

⇒ \(\mathrm{h}=\frac{4}{3}, \mathrm{~g}=3, \mathrm{f}=-\frac{22}{3}, \mathrm{c}=5\)

⇒ \( \mathrm{~g}+\mathrm{c}+\mathrm{h}-\mathrm{f}=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14\)

Hence option(1) is correct

Concept:

Condition for Pair of straight Lines:

The equation ax² + 2hxy + by² = 0 is a homogenous equation of the second degree, representing a pair of straight lines passing through the origin. But
(i) If h² > ab, then the two straight lines are real and different.
(ii) If h² = ab, then the two straight lines are coincident.
(iii) If h² < ab, then the two straight lines are imaginary, having the origin as the point of intersection.

Solution:

(2l − 3)x2 + 2lxy − y2 = 0  . . . (1)

On comparing with  ax² + 2hxy + by² = 0        . . . (2)

we get, a = 2I - 3, h = I, b = - 1

Equation (1) represents a pair of straight lines if 

h² > ab ⇒ I² > (2I - 3)(-1)

⇒ I² >(3 - 2I)

⇒ I² + 2I - 3 > 0

⇒ I² + 3I - I - 3 > 0

⇒ (I + 3)(I - 1) > 0

⇒ I < - 3 or I > 0

∴ I ∈ R - (- 3, 1)

Hence option (2) is correct.